Question
Question: Match the column I with Column II Column I(Physical Quantity)| Column II(Dimensional formula)...
Match the column I with Column II
Column I(Physical Quantity) | Column II(Dimensional formula) |
---|---|
A) Permittivity of free space | p) [M0L0T−1] |
B) Radiant flux | q) [ML3T−3A−2] |
C) Resistivity | r) [ML2T−3] |
D) Hubble constant | s) [M−1L−3T4A2] |
- A-p, B-q, C-r, D-s
- A-q, B-p, C-s, D-r
- A-s, B-r, C-q, D-p
- A-r, B-s, C-q, D-p
Solution
Hint: The dimensions used for the fundamental quantities are as follows
Mass [M]
Length [L]
Time [T]
Current [A]
Complete step by step answer:
The expression for permittivity of free space is given as,
ε0=4πFr2q1q2
Dimensions of force, [F]=[MLT−2]
Dimensions of charge, [q]=[AT]
Dimension of radius, [r]=[L]
Substituting all these dimensions for the dimension of permittivity of free space,
[ε0]=[MLT−2][L2][AT][AT] =[ML3T−2][A2T2] =[M−1L−3T4A2]
Dimension of permittivity of free space, [ε0]=[M−1L−3T4A2]
Radiant flux is the radiant energy transmitted per unit time. This is denoted as ϕe and is measured in Watt. And Watt is the unit of power. Hence the dimension of Radiant flux is the dimension of power.
Power=Force× Velocity
Dimension of Force, [F]=[MLT−2]
Dimension of Velocity,[V]=[LT−1]
Substituting all these dimensions for the dimension of Radiant flux,
Dimension of Radiant flux,
[ϕe]=[MLT−2][LT−1] =[ML2T−3]
Thus, the Dimension of Radiant flux, [ϕe]=[ML2T−3]
The expression for the resistivity is given as,
ρ=lRA
Where,R is the resistance, Ais the area and lis the length.
And the expression for Resistance is given as,
R=IV
Where, V is the voltage and I is the current.
The voltage is the work done per unit charge. Hence the expression for voltage is given as,
V=qW
Therefore, Resistance can be expressed as, R=IqW
Dimension of work done, [W]=[ML2T−2]
Dimension of charge, [q]=[AT]
Dimension of current, [I]=[A]
Substitute all these dimensions for the dimension of resistance.
[R]=[AT][A][ML2T−2] =[ML2T−3A−2]
Dimension of Area, [A]=[L2]
Dimension of length, [l]=[L]
Substitute all these dimensions for the dimension of resistivity,
[ρ]=[L][ML2T−3A−2][L2] =[ML3T−3A−2]
Dimension of resistivity,[ρ]=[ML3T−3A−2]
Hubble constant is the unit of measurement for describing the expansion of the universe.
The expression for Hubble constant is given as,
H = (Velocity/Distance)
Dimension for velocity,[V]=[LT−1]
Dimension for distance,[l]=[L]
Substitute all these dimensions for the dimension of Hubble constant.
[H]=[L][LT−1] =[T−1]
Dimension of Hubble constant,[H]=[M0L0T−1]
So, the correct answer is option 3.
Note:
Dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity. The dimension can be found from the unit of the quantity measured. If any fundamental quantity is absent then the power will be zero in the dimensional formula.